An equilateral triangle is inscribed in the parabola $y^2=16 a x$ with one of its vertices at the origin. Then, the centroid of that triangle is

• A

  $(8 a, 0)$
• B

  $(16 a, 0)$
• Correct answer
  C

  $(32 a, 0)$
• D

  $(48 a, 0)$

Solution

Let the length at one side is $l$, so coordinate of $A$ is $\left(\frac{\sqrt{3} l}{2}, \frac{l}{2}\right)$ lies on parabola so in $y^2=16 a x$ $$ \begin{aligned} \frac{l^2}{4} & =16 a \frac{\sqrt{3}}{2} l \\ \Rightarrow \quad l & =\frac{16 \sqrt{3}}{2} \cdot 4 a \Rightarrow 32 \sqrt{3} a \\ \Rightarrow \quad A & =(48 a, 16 \sqrt{3} a) 0(0,0) B(48 a,-16 \sqrt{3}) \\ \text { Centroid } & =\left(\frac{48 a+48 a+0}{3}, \frac{16 \sqrt{3 a}-16 \sqrt{3 a}+0}{3}\right) \\ & =(32 a, 0) \end{aligned} $$

Hence, the correct answer is option: 'C'.